Let $M$ be a (closed) Riemannian manifold and let $f$ be an isometry of $M$ fixing a point $\ast \in M$ that acts trivially on $\Gamma := \pi_1(M,\ast)$.
Then there is a unique isometry $\tilde{f}$ of the total space of the universal covering $\pi \colon \tilde{M} \to M$ that lifts $f$, i.e., $\pi \circ \tilde{f} = f \circ \pi$, and fixes the set $\pi^{-1}(\ast)$ point-wise. (Note that we can identify this set with $\Gamma$.)
Can it happen that there is a point $x \in \tilde{M}$ and $\gamma \in \Gamma \backslash \{1\}$ such that $\tilde{f}(x) = \gamma(x)$?
If we drop everything pertaining to the metric, such examples indeed exist: for instance we could take $M$ to be a 2-torus so that $\tilde{M} = \mathbf R^2$ and $\tilde{f} \colon (x,y) \mapsto (x+d(y),y)$ with $d\colon \mathbf R \to [0,1]$ smooth, periodic with period $1$, and such that $d(0) = 0, d(1/2) = 1$. But I do not see how this could be modified so that the resulting diffeomorphism $f$ of $M$ is an isometry.
